Problem: Simplify; express your answer in exponential form. Assume $n\neq 0, k\neq 0$. $\dfrac{{(n^{3}k)^{-1}}}{{(n^{-1}k^{-2})^{-5}}}$
To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(n^{3}k)^{-1} = (n^{3})^{-1}(k)^{-1}}$ On the left, we have ${n^{3}}$ to the exponent ${-1}$ . Now ${3 \times -1 = -3}$ , so ${(n^{3})^{-1} = n^{-3}}$ Apply the ideas above to simplify the equation. $\dfrac{{(n^{3}k)^{-1}}}{{(n^{-1}k^{-2})^{-5}}} = \dfrac{{n^{-3}k^{-1}}}{{n^{5}k^{10}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{-3}k^{-1}}}{{n^{5}k^{10}}} = \dfrac{{n^{-3}}}{{n^{5}}} \cdot \dfrac{{k^{-1}}}{{k^{10}}} = n^{{-3} - {5}} \cdot k^{{-1} - {10}} = n^{-8}k^{-11}$